2 should not go out for a walk

3 will quit it at once

4 must buy a smaller one

5 may be very worried

6 may find another one

7 can understand the lecture well

8 has to leave the building in order

9 ought to come to visit me more

10 must be very happy

19

Từ pt đầu ta có:

\(x^2-xy-2xy+2y^2=0\)

\(\Leftrightarrow x\left(x-y\right)-2y\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-2y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\x=2y\end{matrix}\right.\)

TH1: \(x=y\) thế xuống pt dưới:

\(y^2-y-y^2=1\Rightarrow y=-1\Rightarrow x=-1\)

TH2: \(x=2y\) thế xuống pt dưới:

\(\left(2y\right)^2-2y-y^2=1\Leftrightarrow3y^2-2y-1=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-\dfrac{1}{3}\Rightarrow x=-\dfrac{2}{3}\end{matrix}\right.\)

Vậy nghiệm của hệ là: \(\left(x;y\right)=\left(-1;-1\right);\left(1;2\right);\left(-\dfrac{1}{3};-\dfrac{2}{3}\right)\)

21.

Từ pt đầu:

\(xy+2=2x+y\Leftrightarrow xy-y+2-2x=0\)

\(\Leftrightarrow y\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(y-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

TH1: \(x=1\) thế xuống pt dưới:

\(2y+y^2+3y=6\Leftrightarrow y^2+5y-6=0\)

\(\Rightarrow\left[{}\begin{matrix}y=1\\y=-6\end{matrix}\right.\)

TH2: \(y=2\) thế xuông pt dưới

\(4x+4+6=6\Rightarrow x=-1\)

Vậy nghiệm của pt là: \(\left(x;y\right)=\left(1;1\right);\left(1;-6\right);\left(-1;2\right)\)

Ai giúp em với ạ

chữ đẹp quá :))

Chữ thầy e á a😥😥

Bài 6:

Xét ΔOAC vuông tại A và ΔOBD vuông tại B có

OA=OB

\(\widehat{AOC}=\widehat{BOD}\)(hai góc đối đỉnh)

Do đó: ΔOAC=ΔOBD

=>OC=OD

Bài 7:

a: Ta có: \(\widehat{DAB}+\widehat{BAC}+\widehat{CAE}=180^0\)

=>\(\widehat{DAB}+\widehat{CAE}+90^0=180^0\)

=>\(\widehat{DAB}+\widehat{CAE}=90^0\)

mà \(\widehat{DAB}+\widehat{DBA}=90^0\)

nên \(\widehat{DBA}=\widehat{CAE}\)

Xét ΔABD vuông tại A và D và ΔCAE vuông tại E có

AB=AC

\(\widehat{DBA}=\widehat{EAC}\)

Do đó: ΔABD=ΔCAE

b: ta có: ΔABD=ΔCAE

=>DB=AE và AD=CE

DB+CE=DA+AE=DE

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a: Ta có: \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Leftrightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Leftrightarrow-9x=-18\)

hay x=2

b: ta có: \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+x-6x^2+5x-3=0\)

\(\Leftrightarrow17x=-5\)

hay \(x=-\dfrac{5}{17}\)

c: Ta có: \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2-12-14x+13=0\)

\(\Leftrightarrow8x^3-8x=0\)

\(\Leftrightarrow8x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

a) \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Rightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Rightarrow-9x=-18\)

\(\Rightarrow x=2\)

b) \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Rightarrow x^3+6x^2+12x+8-x^3+x=6x^2-5x+3\)

\(\Rightarrow18x=-5\)

\(\Rightarrow x=-\dfrac{5}{18}\)

c) \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Rightarrow8x^3-12x^2+6x-1+12x^2-12=14x-13\)

\(\Rightarrow8x^3-8x=0\)

\(\Rightarrow8x\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)